While explaining the Fermat’s Theorem, Prateek Bhaiya took an example of finding A^B. It’s clear upto the point where he asks to represent B in terms of p. But at 5:04, he takes mod p with entire A^B. It’s not clear to me why has he taken A^B % p.
Why was mod taken with A^B
hello @pawan2179
we know x^(p-1) mod p => 1 if p is prime number…fermant little theorem
now to calculate (a^b) mod p .
we can write b = (p-1) + (p-1) + (p-1) … + k
where k is remainder we are left with. i.e k=b%(p-1)
substituting value of b
(a^b)mod p = a^(p-1 + p-1 + p-1 + p-1 … + k) mod p
now using property a^(x+y) => a^x * a^y
=> ( a^(p-1) * a^(p-1) * a^(p-1) … a^k )mod p
now we know a^(p-1) mod p =1
=> 1 * 1 * 1 …a^k mod p
=> a^k mod p
substituting value of k
=> a^(b%(p-1) ) mod p
I hope I’ve cleared your doubt. I ask you to please rate your experience here
Your feedback is very important. It helps us improve our platform and hence provide you
the learning experience you deserve.
On the off chance, you still have some questions or not find the answers satisfactory, you may reopen
the doubt.